a********l
发帖数: 55 | 1 考虑一个measurable space (X,B)，其中X是[0,1]，B是collection of Borel sets on
[0,1].
在(X,B)上，所有measure都可以用cdf来表达。（是吗？）
如果我有一个sequence of measures {m_n} 和一个measure m，请问我怎样用cdf来保
证{m_n}是setwise convergent to m 呢？cdf 是pointwise convergent可以吗？如果
不行，cdf是uniformly convergent可以吗？
先谢过！ | a********l
发帖数: 55 | 2 补充：
m_n 和 m 是 measures, 所以是 set functions.
根据Royden "Real Analysis"上的定义：
m_n converges setwise to m if for each E in B we have m(E) = lim m_n(E). | Q***5
发帖数: 994 | 3 I guess the condition that 'cdf pointwise convergence' does not guarentee
the setwise convergence. Here is a counter example:
Let m has point mass at 1/2, with prob = 1, so the cdf of m is step function
with value 0 on [0 1/2) and 1 on [1/2, 1]
Let m_n be measures with cdf of (2x)^n on [0,1/2) and 1 on [1/2,1]. | a********l
发帖数: 55 | 4 Thank you very much! You are right!
Then do you think uniform convergence of cdf is sufficient? Thanks. | Q***5
发帖数: 994 | 5 No. Example:
Let m be the flat distribution on [0,1], the cdf is y=x
For each n, divide [0,1] evenly into n segments and define m_n to have point
mass probability 1/(n+1) at end points and 0 otherwise.
Then the cdf of m_n is a piecewise constant function, which uniformly
converges to y=x, but m_n do not set wise converges to m on Q, the set of
all ratinonal numbers on [0,1].
【在 a********l 的大作中提到】
: Thank you very much! You are right!
: Then do you think uniform convergence of cdf is sufficient? Thanks.
| a********l
发帖数: 55 | 6 Thanks. Again you are right.
But what if I also assume that all the cdf's associated with {m_n} and m are
an equicontinuous family of functions? Say, they are all Lipschitz
continuous with Lipschitz constants not larger than 1.
Thanks a lot!!!
point
【在 Q***5 的大作中提到】
: No. Example:
: Let m be the flat distribution on [0,1], the cdf is y=x
: For each n, divide [0,1] evenly into n segments and define m_n to have point
: mass probability 1/(n+1) at end points and 0 otherwise.
: Then the cdf of m_n is a piecewise constant function, which uniformly
: converges to y=x, but m_n do not set wise converges to m on Q, the set of
: all ratinonal numbers on [0,1].
| Q***5
发帖数: 994 | 7 The assumption of bounded Lipschitz constant B (B should be something
greater than 1 ) and point wise convergence of the cdf's will be
sufficient.
Sketch of proof: ( Let p be the flat distribution on [0 1] )
(1) For any measurable set A, we can show that m_n(A)<= B p(A) and m(A)<=B p
(A)
(2) To prove by contradiction, we can assume (without loss of generality)
that there is a Borel-measurable set D, such that m_n(D)>m(D)+delta,for all
n. (delta>0 is a constant). Then by (1), we can find a open
【在 a********l 的大作中提到】
: Thanks. Again you are right.
: But what if I also assume that all the cdf's associated with {m_n} and m are
: an equicontinuous family of functions? Say, they are all Lipschitz
: continuous with Lipschitz constants not larger than 1.
: Thanks a lot!!!
:
: point
| Q***5
发帖数: 994 | 8 I mean "p(E\F) is so small " | H****h
发帖数: 1037 | 9 cdf逐点收敛等价于weak convergence.
on
【在 a********l 的大作中提到】
: 考虑一个measurable space (X,B)，其中X是[0,1]，B是collection of Borel sets on
: [0,1].
: 在(X,B)上，所有measure都可以用cdf来表达。（是吗？）
: 如果我有一个sequence of measures {m_n} 和一个measure m，请问我怎样用cdf来保
: 证{m_n}是setwise convergent to m 呢？cdf 是pointwise convergent可以吗？如果
: 不行，cdf是uniformly convergent可以吗？
: 先谢过！
| a********l
发帖数: 55 | 10 Thanks so much for the sketch of the proof, QL.
I think it's correct. I just have a problem to find a rigorous proof for
your step (1), although it is "obviously" true. Could you please give me
some hints, or a reference?
Actually, does this sufficient condition for setwise convergence of measure
appear in books, so that I can quote.
Thanks.
p
all
union
【在 Q***5 的大作中提到】
: The assumption of bounded Lipschitz constant B (B should be something
: greater than 1 ) and point wise convergence of the cdf's will be
: sufficient.
: Sketch of proof: ( Let p be the flat distribution on [0 1] )
: (1) For any measurable set A, we can show that m_n(A)<= B p(A) and m(A)<=B p
: (A)
: (2) To prove by contradiction, we can assume (without loss of generality)
: that there is a Borel-measurable set D, such that m_n(D)>m(D)+delta,for all
: n. (delta>0 is a constant). Then by (1), we can find a open
| | | Q***5
发帖数: 994 | 11 I think you can prove (1) by the following arguments:
1.1 It is true when A is any open segment -- This is just by defination of
cdf.
1.2 It is true for any open subset of [0 1]. An open subset is just a
uninon of countably many opens segments.
1.3 It is true for any Borel measurable subset A: m_n(A) = inf_{A\subset D,
D open}m_n(D)<= B\inf_{A\subset D, D open}m(D) = B m(A)
measure
【在 a********l 的大作中提到】
: Thanks so much for the sketch of the proof, QL.
: I think it's correct. I just have a problem to find a rigorous proof for
: your step (1), although it is "obviously" true. Could you please give me
: some hints, or a reference?
: Actually, does this sufficient condition for setwise convergence of measure
: appear in books, so that I can quote.
: Thanks.
:
: p
: all
| a********l
发帖数: 55 | 12 Thanks QL. I still have a problem for your step 1.3. I don't know how to
prove that m_n(A) = inf_{A\subset D, D open}m_n(D).
In other words (using Royden's terminology), how can I prove that the
measure m_n is outer regular for A?
Many thanks!
of
D,
【在 Q***5 的大作中提到】
: I think you can prove (1) by the following arguments:
: 1.1 It is true when A is any open segment -- This is just by defination of
: cdf.
: 1.2 It is true for any open subset of [0 1]. An open subset is just a
: uninon of countably many opens segments.
: 1.3 It is true for any Borel measurable subset A: m_n(A) = inf_{A\subset D,
: D open}m_n(D)<= B\inf_{A\subset D, D open}m(D) = B m(A)
:
: measure
| Q***5
发帖数: 994 | 13 Should that be the definition?
My knowledge of measure theory is a little rusty now. Can you remind me how
do you define the measure on a Borel measurable set given the cdf?
【在 a********l 的大作中提到】
: Thanks QL. I still have a problem for your step 1.3. I don't know how to
: prove that m_n(A) = inf_{A\subset D, D open}m_n(D).
: In other words (using Royden's terminology), how can I prove that the
: measure m_n is outer regular for A?
: Many thanks!
:
: of
: D,
| Q***5
发帖数: 994 | 14 For (1.3), you can pull a big gun: Radon-Nikodym Thm.
We know that m(A)< B p(A) for any open set. So for any Borel measuable set D
, if p(D) = 0, then D can be covered by a open set, which is of arbitrarily
small measure. As a result, m(D) is arbitarily small, there for m(D) = 0.
Hence m<=0, and h is Borel
measurable, L_1 w.r.t p.
You can prove that h is bounded by B a.e. w.r.t p: otherwise, there exists
delta>0, such that p(E)>0, where E = (h>B+delta), w | a********l
发帖数: 55 | 15 You are right. Thank you so so so much!
D
arbitrarily
that
【在 Q***5 的大作中提到】
: For (1.3), you can pull a big gun: Radon-Nikodym Thm.
: We know that m(A)< B p(A) for any open set. So for any Borel measuable set D
: , if p(D) = 0, then D can be covered by a open set, which is of arbitrarily
: small measure. As a result, m(D) is arbitarily small, there for m(D) = 0.
: Hence m<=0, and h is Borel
: measurable, L_1 w.r.t p.
: You can prove that h is bounded by B a.e. w.r.t p: otherwise, there exists
: delta>0, such that p(E)>0, where E = (h>B+delta), w
|
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